Duct Design
MSYS4480
Duct design
1. Air flow in ducts
2. Major and Minor Losses in Ducts
3. Loss coefficient for some fittings
4. Equivalent length for a fittings
5. Duct accessories
6. Pressure diagram
7. Duct design
1. Equal friction method
2. Balanced Capacity method
8. Flex Ducts
9. In-Slab Ducts
10. Avoiding Bullhead Tees
11. Return Air Boots
12. Pressurized Plenums with Home Run Ducts
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Air Flow in Ducts
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Internal, External, and Total Static Pressure Drop
Internal Static Pressure losses occur External Static Pressure (ESP) losses Total Static Pressure (TSP) loss is the
within mechanical equipment and are occur within the system outside of the sum of the internal and external losses
usually calculated by the manufacturer mechanical equipment and are usually in the system.
Examples include calculated by the mechanical
consultant. Examples include
▪ Dampers ▪ Louvers
▪ Filters ▪ Dampers (motorized, balancing,
backdraft…)
▪ Coils
▪ Duct fittings
▪ Heat exchangers
▪ Duct transitions and elbows
▪ Heat recovery devices (such as
wheels, heat pipes) ▪ Air terminals
▪ Air valves and VAV boxes
▪ Filters
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Steady Flow Equation
Volume flow rate = Area x
air velocity
A1 x V1 = A2 x V2
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Example 1: Calculating Air Flow
An airfow velocity through a duct with an area of 1ft² is 1,000 fpm.
Determine the new airflow velocity when the area is increased to 4 ft².
A1 x V1
V2
A2
1 ²x
V2 = 500 fpm
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Steady Flow Energy Equation
Ps1 + PV1 + Pe1 + Pp – Pf = Ps2 + PV2 + Pe2
Ps1 + PV1 + Pp – Pf = Ps2 + PV2 + Pe2
Ps1 + (V12/2g) + Pp – Pf = Ps2 + (V22/2g)
Ps1 = Static Pressure of air at section 1 [ft]
V12 = Velocity at section 1 [ft/sec]
Pe1 = Elevations at section 1 [ft]
Pp = Pressure added by the fan [ft]
G = gravitational constant [32.2 ft/sec²]
Pf = Pressure loss in duct by friction [ft]
Ps2 = Static Pressure of air at section 2 [ft]
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Example 2: Calculating Static Pressure Drop
Given the Volume Flow Rate through a duct is 8,000 cfm, the friction loss from
point 1 to point 2 is 0.5” WC. and the static pressure at point 1 is 2” WC.
Q1 = 8,000 cfm Q1 = 8,000 cfm
A1 = 4ft² A1 = 16ft²
P1 = 2” WC P1 = ?
Ps1 + (V12/2g) + Pe1 + Pp – Pf = Ps2 + (V22/2g) + Pe2
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Velocity air pressure, Pv
Pv in in water and V in ft/min
Pv in Pa and V in m/s
Mass Density ρ
62.4 lbm/ft3 and 999 kg/
Alternate units: Ft
²
PVelocity [ft], V = ft/sec Pressure changes during flow in ducts.
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TOTAL Pressure
PTotal = PVelocity + PStatic
cannabis
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Friction Loss
▪ Tedious task to solve by equations
▪ Pressure Loss Charts have been prepared.
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Friction Loss
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Friction Loss
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Equivalent of a circular duct
Dh = Hydraulic diameter
a and b are the dimension of a rectangular duct
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Equivalent of a circular duct
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Example 3: Calculate Pressure Loss
Compute the lost pressure in a 6 in., 90-degree pleated elbow that has 150 cfm of air flowing through it. The ratio of
turning radius to diameter is 1.5. Assume standard air.
Table 12-8: the loss coefficient: 0.43
1” Water Guage = 248.84 Pa
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Friction Loss Fitting Table
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Friction Loss Fitting Table
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Friction Loss Fitting Table
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Friction Loss Fitting Table
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Friction Loss Fitting Table
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Friction Loss Fitting Table
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Example 4: Pressure Loss
Compute the loss in total pressure
for a round 90-degree branch and
straight-through section, a tee.
The common section is 12 in. in
diameter, and the straight-through
section has a 10 in. diameter with a
flow rate of 1100 cfm.
The branch flow rate is 250 cfm
through a 6 in. duct.
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Equivalent lengths
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Example 5: Friction Loss Example
Compute the equivalent lengths for the fittings in the duct system below.
The fittings are an entrance, a 45-degree wye, the straight-through
section of the wye fitting, a 45-degree elbow, and a 90-degree elbow.
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Friction Loss Example
Example:
What is the total pressure loss on the critical
path?
Critical Path: 1-a-3 with Equivalent length
of 127.4 ft
Path 1: 1-a-2
Le = Li + 50 + Lwye,S + 50 We pick average pressure (friction) loss for
= 19 + 50 + 4.4 + 50 duct and calculate the total pressure loss
= 123.4 ft for the system.
Path 2: 1-a-3 Friction loss to be designed for =
Le = Li + 50 + Lwye,Br + Lelbow-90 + 40 0.08”/100ft
= 19 + 50 + 11 + 7.7 + 40
= 127.4 ft P = 127.4 ft x 0.08”/100ft = 0.102”
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Duct Accessories
1. Turning vanes
▪ Linear
▪ Airfoil (More efficient)
2. Dampers
▪ Parallel blades (open/close)
▪ Opposed blades (modulate airflow)
▪ Balancing
▪ Motorized
▪ Backdraft
3. Fire dampers
▪ Type A (blades inside air stream)
▪ Type B (blades outside air stream)
4. Electric duct heaters
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Turning Vanes
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Dampers
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Air Flow in Ducts
Volume (Q) is a function of cross
sectional
area (A) and velocity (V)
Q=AV
however, momentum, friction and
turbulence must also be accounted for in
the sizing method
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Static Pressure
▪ Force required to overcome friction and loss of momentum due to turbulence
▪ As air encounters friction or turbulence, static pressure is reduced
▪ Fans add static pressure
▪ Static pressure is measured in Inches-water gauge
▪ Positive pressure pushes air
▪ Negative pressure draws air
▪ Straight ducts have a pressure loss of “w.g./100’
based on diameter and velocity
Equivalent Length
▪ Describes the amount of static pressure lost in a fitting that would be comparable to a length of straight duct
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Duct Construction
▪ Round ductwork is the most efficient but requires greater depth
▪ Rectangular ductwork is the least efficient but can be reduced in depth to accommodate smaller
clearances
▪ Avoid aspect ratios greater than 5:1
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Flex Ducts
▪ Used to dampen noise when connecting to air terminals or mechanical equipment (i.e. bathroom fans)
▪ Typically only used for a max 5 foot length.
▪ Long runs of flex duct and elbows create large pressure drops in your system.
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In-Slab Ducts
▪ Typically seen in high rise buildings were no dropped ceilings are given near building exterior.
▪ Used to vent oven ranges, dryers, and sometimes bathroom fans.
▪ Can handle little airflow (approx. 50 CFM) due to size.
▪ Elbows always shown as two 45° joints to minimize pressure drop.
▪ Must be minimum of 2’-0” from structural bearing entities (columns, walls).
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No “Bull Head” Tees
Airflow does not travel well when there is no clear path to
follow. Instead
▪ show the duct continuing onward past the branch (as shown
below),
▪ add turning vanes.
▪ use “pant leg” or wye type fitting (best option but most
expensive)
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Return Air Boot
Used to reduce noise emanating from
mechanical equipment in and adjacent
to occupied spaces.
Should completely obscure line of sight
to the air inlet. This forces the sound to
bounce
Specified with 1” acoustic insulation.
Typical “L” shaped boot is shown to the
right.
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Z-Shape Return Air Boot
Less common but more effective.
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Pressurized Plenum with Home Run Ducting
Can be used where there are multiple duct
diffusers with similar airflow requirements.
Each “home run” should be approximately the
same length with the same pressure drop.
Do not take off ducts close to fan coil or at end
of plenum.
Advantages:
▪ Can be used where there is very little ceiling
height (i.e. running four 6”ø ducts as
opposed to one 10”ø / 10”x8”)
▪ Requires less overall space
Disadvantages
▪ Cannot handle large pressure drops
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Questions?
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Equal Friction Method
▪ Presumes that friction in ductwork can be balanced to allow uniform friction loss through all
branches
1. Find effective length (EL) of longest run
2. Establish allowed static pressure loss/100’
∆P=100(SP)/EL
3. Size ducts
4. Repeat for each branch
Note: velocity must be higher in each upstream section
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Assignment #2: Equal Friction Method
Given:
• The system shown is supplied air by a rooftop
unit that develops 0.25 in. wg total pressure
external to the unit.
• The return air system requires 0.10 in. wg.
• The ducts are to be of round cross section,
and the maximum velocity in the main run is
850 ft/min, whereas the branch velocities
must not exceed 650 ft/min.
Size:
• The ducts using the equal-friction method.
• Show the location of any required dampers.
Compute the total pressure loss for the
system.
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Equal Friction Method - Example
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